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 2005-08-29, 19:44 #1 Xyzzy     "Mike" Aug 2002 22×2,011 Posts Math rendering engine $\2^{57,885,161}-1$ This is using our new $\small \TeX$ tags... Look at the last entry here: http://www.mersenneforum.org/misc.php?do=bbcode Manual: http://www.mersenneforum.org/mimetex.htm
 2005-08-29, 20:06 #2 paulunderwood     Sep 2002 Database er0rr 70428 Posts Great!! now we can write things like $e^{i\pi}+1=0$ and $\sum_{i=1}^{n}x_i=\int_0^{1}f$. We need a few good links to online tex guides...
2005-08-29, 20:08   #3
xilman
Bamboozled!

"𒉺𒌌𒇷𒆷𒀭"
May 2003
Down not across

2·5,323 Posts

Quote:
 Originally Posted by Xyzzy $\2^p-1$ This is using our new $\small \TeX$ tags... Look at the last entry here: http://www.mersenneforum.org/misc.php?do=bbcode
Excellent!

I've long wanted to be able to use $\TeX$ formatting.

Paul

 2005-08-29, 20:18 #4 Xyzzy     "Mike" Aug 2002 22×2,011 Posts There are several reasons why this took so long to get set up. The first reason is our old server didn't run Linux so a lot of software wasn't available. Secondly, we had very limited access to the server which prevented us from doing really cool stuff. We were also was hesitant to dive into a solution that didn't have a workable license. As it is, it pains us that the forum software itself isn't GPLed. We have no idea how much of a load this will put on the server, but at any time we can redirect the image generation to another box running this. For now, especially if you are going to test some really crazy stuff, use the interactive editor thing. Now that we have this set up, we realize we could have had this a long time ago, by redirecting the image generation to another box. We apologize that it took us this long to figure it out. Who is going to post the most interesting example formula? Have fun!
 2005-08-29, 23:30 #5 Numbers     Jun 2005 Near Beetlegeuse 22·97 Posts Just testing. $\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ Could be bigger, don't you think? Last fiddled with by Numbers on 2005-08-29 at 23:31
2005-08-29, 23:39   #6
Xyzzy

"Mike"
Aug 2002

1F6C16 Posts

Quote:
 Originally Posted by Numbers Could be bigger, don't you think?
$\LARGE \displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

 2005-08-30, 00:00 #7 Numbers     Jun 2005 Near Beetlegeuse 22·97 Posts $\displaystyle2\left(\frac12+\frac23\right)=\frac53$ Xyzzy, I think you're an absolute genius. But you're going to have to tell me how to make it bigger, because even in displaystyle it still comes out pretty small.
2005-08-30, 00:03   #8
Xyzzy

"Mike"
Aug 2002

22·2,011 Posts

Quote:
 Originally Posted by Numbers But you're going to have to tell me how to make it bigger, because even in displaystyle it still comes out pretty small.
Just hit "Quote" to see the source and you will see we added a \LARGE command. The documentation covers this. We have a sneaky feeling we could spend a lot of time learning all this. It seems to be very powerful. The interactive editor we linked to earlier makes it easier to try new stuff quickly.

2005-08-30, 00:14   #9
Wacky

Jun 2003
The Texas Hill Country

100010000012 Posts

Quote:
 Originally Posted by Numbers $\displaystyle2\left(\frac12+\frac23\right)=\frac53$
Big, or small, I don't like the math.
$\LARGE\displaystyle2\left(\frac12+\frac23\right)=\frac73$

 2005-08-30, 02:04 #10 jinydu     Dec 2003 Hopefully Near M48 2·3·293 Posts So there is more than one type of Latex? $\displaystyle{f(x)=f(a)+f'(a)x+\frac{1}{2}f''(a)x^2+\frac{1}{6}f'''(a)x^3+...}$ (I guess that answers my question. What I just typed would surely have worked on a certain other forum). Last fiddled with by jinydu on 2005-08-30 at 02:05
 2005-08-30, 04:32 #11 Dougy     Aug 2004 Melbourne, Australia 23·19 Posts Often I use congruences, so I'll just pick a theorem at random... Theorem: Let $p$ be a prime such that $p^2$ divides $M_q=2^q-1$ for some prime $q$. Then $2^{\frac{p-1}{2}} \equiv 1 \pmod {p^2}$ and so $p$ is a Wieferich prime. Proof: We know that if $p$ divides $M_q=2^q-1$, then $p=2kq+1$ for some $k \in \mathbb{N}$. Then $q=\frac{p-1}{2k}$, and if $p^2$ divides $M_q=2^q-1$ then $2^q \equiv 2^{\frac{p-1}{2k}} \equiv 1 \pmod {p^2}$.

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